Hamiltonian Mechanics and Completely Integrable Systems

Since $d\omega =0$, [F2] reduces to ${\mathcal{L}}_X\omega =d({\iota }_X\omega )$.

Therefore the left side vanishes exactly when ${\iota }_X\omega$ is closed, which is precisely the equivalence in [F1].

Nondegeneracy says that ${\omega }^{\mathrm{♭}}:TM\to T^{\ast }M$ is a fibrewise linear isomorphism, and its local matrix and inverse are smooth.

Thus $X_H=({\omega }^{\mathrm{♭}}{)}^{-1}(dH)$ is smooth, satisfies [F1], and is the only possible solution because ${\omega }^{\mathrm{♭}}$ is injective.

If $X=X_H$, then ${\iota }_X\omega =dH$ is exact and therefore closed; [F1] makes $X$ symplectic.

If $X$ is symplectic, [F1] makes ${\iota }_X\omega$ closed. Around any point restrict to a coordinate ball that is star-shaped in coordinates. The coefficient vector of this smooth one-form satisfies the symmetric-partial equations for a closed field, so [F2] supplies $H$ there with $dH={\iota }_X\omega$. Thus $X=X_H$ locally.

If both functions generate $X$, [F1] gives $d(H-K)=0$. In a connected coordinate ball, integration along line segments shows that a smooth function with zero differential is constant; hence $H-K$ is locally constant and therefore constant on each connected component.

Conversely, if $c$ is locally constant then $dc=0$, so $d(H+c)=dH$ and [F1] gives $X_{H+c}=X_H$.

The linear bundle isomorphism ${\omega }^{\mathrm{♭}}$ gives a linear bijection between all vector fields and all one-forms. By [F1] it restricts to a bijection from symplectic fields to closed one-forms.

By [F2], the inverse image of the exact one-forms is precisely the Hamiltonian fields. Passing to quotients in step 1.1 therefore gives $Z^1\mathrm{/}d{C}^{\mathrm{\infty }}=H_{\mathrm{d}\mathrm{R}}^1(M)$ and the displayed natural isomorphism.

Since ${\iota }_{X_H}\omega =dH$ is closed, [F1] gives ${\mathcal{L}}_{X_H}\omega =0$.

Apply [F2] on the domain of the local flow to obtain ${\varphi }_t^{\ast }\omega =\omega$. Neither step extends the flow beyond its maximal domain.

Along $\gamma$, $\frac{d}{dt}H(\gamma (t))=dH(X_H)=\omega (X_H,X_H)=0$ by alternation.

Hence $H\circ \gamma$ is constant on every connected time interval in the maximal domain. This proves conservation without assuming completeness.

Linearity of $d$ and uniqueness of Hamiltonian fields give $X_{aF+bG}=a{X}_F+b{X}_G$. Bilinearity and alternation of $\omega$ now make the bracket bilinear and skew.

Since a vector field is a derivation, [F1] gives $\{FG,H\}=X_H(FG)=F\{G,H\}+G\{F,H\}$. Skew-symmetry then gives the displayed Leibniz rule in the second entry as well.

Apply [F2] to $\omega$: ${\iota }_{[X_F,X_G]}\omega ={\mathcal{L}}_{X_F}(dG)-{\iota }_{X_G}({\mathcal{L}}_{X_F}\omega )=d(X_{F}G)=-d\{F,G\}\mathrm{.}$

The right side is ${\iota }_{-X_{\{F,G\}}}\omega$. Nondegeneracy makes contraction injective, so the vector fields are equal.

Expand $0=d\omega (X_F,X_G,X_H)$. Replacing derivatives by $X_A(B)=\{B,A\}$ and commutators by [F1], the six terms combine in equal pairs to $0=2(\{\{G,H\},F\}+\{\{H,F\},G\}+\{\{F,G\},H\})\mathrm{.}$ This is a pointwise identity, not merely a statement that its differential vanishes.

Divide by two and use skew-symmetry on each outer bracket. The result is the displayed Jacobi identity.

Pointwise addition and multiplication make $C^{\mathrm{\infty }}(M)$ a commutative associative real algebra with unit, and [F1] supplies a bilinear skew biderivation.

By [F2] that bracket is a Lie bracket. These are exactly the Poisson-algebra axioms, so the claimed structure follows.

The chain rule and $\dot{\gamma }=X_H\circ \gamma$ give $\frac{d}{dt}(F\circ \gamma )=dF(X_H)\circ \gamma$.

By [F1], $dF(X_H)=X_H(F)=\{F,H\}$, proving the formula on the entire local domain of the curve.

If $\{F,H\}=0$, [F1] makes the derivative of $F$ along every integral curve zero. Ordinary one-variable calculus makes $F$ constant on each interval domain, so it is a first integral by [F2].

Conversely, through every $p\in M$ there is a local integral curve. If $F$ is a first integral, its derivative at time zero is zero; [F1] identifies it with $\{F,H\}(p)$. Since $p$ was arbitrary, the bracket vanishes everywhere.

By [F2], the flows commute exactly when $[X_F,X_G]=0$. By [F1], this is equivalent to $X_{\{F,G\}}=0$.

By [F3], the latter condition holds exactly when $\{F,G\}$ is locally constant. The zero bracket is one such function.

Write $X_H={\sum }_i(a^i{\mathrm{\partial }}_{q^i}+b_i{\mathrm{\partial }}_{p_i})$. Then [F1] gives ${\iota }_{X_H}\omega ={\sum }_i(a^{i}d{p}_i-b_{i}d{q}^i)$.

Comparing with $dH={\sum }_i(H_{q^i}d{q}^i+H_{p_i}d{p}_i)$ in [F2] gives $a^i=H_{p_i}$ and $b_i=-H_{q^i}$. Since an integral curve has velocity $X_H$, these are Hamilton's equations.

Apply the vector field in [F1] to $F$: $X_G(F)={\sum }_i(G_{p_i}{F}_{q^i}-G_{q^i}{F}_{p_i})$.

By [F2] this is $\{F,G\}$, and commuting scalar factors gives the displayed formula with the library sign.

In coordinates $Y=Y^i(q){\mathrm{\partial }}_{q^i}$, differentiation of the inverse-transpose lift gives $Y^{\mathrm{\#}}=Y^i{\mathrm{\partial }}_{q^i}-p_j({\mathrm{\partial }}_{q^i}{Y}^j){\mathrm{\partial }}_{p_i}$. Also $H_Y=p_j{Y}^j$.

Contracting with ${\omega }_{\mathrm{c}\mathrm{a}\mathrm{n}}={\sum }_{i}d{q}^i\wedge d{p}_i$ gives ${\iota }_{Y^{\mathrm{\#}}}{\omega }_{\mathrm{c}\mathrm{a}\mathrm{n}}=Y^{i}d{p}_i+p_j({\mathrm{\partial }}_{q^i}{Y}^j)d{q}^i=d(p_j{Y}^j)=d{H}_Y$. By [F2], $Y^{\mathrm{\#}}=X_{H_Y}$.

Cartan's formula and [F1] give ${\mathcal{L}}_{X_{H_t}}\omega =d(d{H}_t)+{\iota }_{X_{H_t}}d\omega =0$.

By [F2], ${\mathrm{\partial }}_t{\mathrm{\Phi }}_{t,s}^{\ast }\omega =0$. At $t=s$ the pullback is $\omega$, so it remains $\omega$ on the evolution domain.

Pullback respects wedges and scalar multiplication, so [F2] gives ${\varphi }_t^{\ast }\mathrm{\Omega }=({\varphi }_t^{\ast }\omega {)}^n\mathrm{/}n!={\omega }^n\mathrm{/}n!$.

Thus ${\varphi }_t^{\ast }\mathrm{\Omega }=\mathrm{\Omega }$ wherever the local flow exists. This is volume preservation, with no completeness conclusion.

Differentiate the identity ${\varphi }_t^{\ast }\mathrm{\Omega }=\mathrm{\Omega }$ from [F1] at $t=0$ to obtain ${\mathcal{L}}_{X_H}\mathrm{\Omega }=0$.

By [F2], $({\mathrm{div}}_{\mathrm{\Omega }}{X}_H)\mathrm{\Omega }=0$. Since $\mathrm{\Omega }$ is nowhere zero, the divergence function vanishes.

For the displayed radial field, contraction with [F1] gives ${\iota }_Z{\omega }_{\mathrm{c}\mathrm{a}\mathrm{n}}=-{\sum }_i{p}_{i}d{q}^i=-\lambda$.

Nondegeneracy makes the solution of [F2] unique, so this radial field is the canonical Liouville field. Its local flow is $(q,p)\mapsto (q,e^{t}p)$.

Restrict $T={\varphi }_{\tau }$ and the symplectic volume measure to $R$. Invariance keeps $T$ on $R$, [F1] makes it measure preserving, and the hypothesis gives finite total measure.

Apply [F2] to $(R,T)$. Since $T^n={\varphi }_{n\tau }$ wherever the iterates are defined, its conclusion is exactly the stated recurrence. No assertion is made for an incomplete time map.

On a chart subinterval, a variation field ${\eta }^i(t)={\mathrm{\partial }}_s{q}_s^i{\mathrm{\mid }}_{s=0}$ with zero endpoint values gives, by differentiation under the finite integral, $\delta {\mathcal{S}}_L={\int }_a^b(L_{q^i}{\eta }^i+L_{v^i}{\dot{\eta }}^i)dt\mathrm{.}$

Apply [F2]. The boundary term $[L_{v^i}{\eta }^i{]}_a^b$ vanishes, leaving $\delta {\mathcal{S}}_L={\int }_a^b(L_{q^i}-\frac{d}{dt}{L}_{v^i}){\eta }^{i}dt$. Thus the displayed equations imply stationarity.

Conversely, if one continuous coefficient $E_i=L_{q^i}-\frac{d}{dt}{L}_{v^i}$ were nonzero at an interior time, it would retain one strict sign on a smaller interval. Choosing a nonnegative smooth bump ${\eta }^i$ supported there and all other components zero would make the integral in step 2.1 nonzero, contradicting stationarity. Hence all $E_i$ vanish. Variations supported in chart subintervals cover the curve, proving the coordinate-independent equivalence.

Differentiate the formula in [F2]. Since $p_i=L_{v^i}(q,v)$, the $p_{i}d{v}^i$ and $L_{v^i}d{v}^i$ terms cancel, giving $dH=v^{i}d{p}_i-L_{q^i}d{q}^i$. Hence $H_{p_i}=v^i$ and $H_{q^i}=-L_{q^i}$.

If $q(t)$ satisfies [F1] and $p(t)=L_v(q(t),\dot{q}(t))$, then step 1.1 gives ${\dot{q}}^i=v^i=H_{p_i}$ and ${\dot{p}}_i=\frac{d}{dt}{L}_{v^i}=L_{q^i}=-H_{q^i}$. Thus $(q,p)$ satisfies [F3].

Conversely, a Hamiltonian trajectory satisfies $\dot{q}=v(q,p)$ by [F3] and step 1.1, so inverse Legendre gives $p=L_v(q,\dot{q})$. Its second Hamilton equation then reads $\frac{d}{dt}{L}_{v^i}=L_{q^i}$, which is [F1]. Hyperregularity makes both assignments global inverses.

Fibre differentiation gives $\mathbb{F}L(q,v)=g_q(v,\cdot )=g_q^{\mathrm{♭}}(v)$. Positive definiteness makes $g^{\mathrm{♭}}$ a smooth bundle isomorphism with inverse $g^{\mathrm{♯}}$, so $L$ is hyperregular.

With $p=g^{\mathrm{♭}}v$, [F1] gives $E_L=g(v,v)-\frac{1}{2}g(v,v)+V=\frac{1}{2}g(v,v)+V$. Substituting $v=g^{\mathrm{♯}}p$ yields $H(q,p)=\frac{1}{2}{g}^{-1}(p,p)+V(q)$.

By [F1], $N$ has dimension $2n-n=n$ and $T_{p}N={\bigcap }_j\ker d{F}_j$. For every $i,j$, $d{F}_j(X_{F_i})=\omega (X_{F_j},X_{F_i})=-\{F_j,F_i\}=0$, so each $X_{F_i}(p)$ is tangent.

The bundle isomorphism ${\omega }^{\mathrm{♭}}$ sends $X_{F_i}$ to $d{F}_i$. Since $N$ is a regular fibre, [F1] says $d{F}_p$ has rank $n$, so these $n$ vectors are independent. By dimension they span $T_{p}N$. Their mutual symplectic pairings are the zero brackets $\{F_i,F_j\}$ from [F2], so $T_{p}N$ is isotropic.

Apply [F3] at every point: the $n$-dimensional isotropic tangent spaces are Lagrangian. Thus $N$ is a Lagrangian submanifold and the displayed spanning formula holds.

Let ${\varphi }_i^t$ be the local flow of $X_{F_i}$. Pairwise involution in complete integrability and [F1] make these flows commute. Therefore $(t_1,\dots ,t_n)\cdot p={\varphi }_1^{t_1}\circ \cdots \circ {\varphi }_n^{t_n}(p)$ is independent of the order and satisfies the action law wherever both sides are defined.

By [F2], every regular fibre is invariant under all these flows. On a compact fibre, a maximal trajectory of any restricted smooth field cannot escape in finite time: a convergent subsequence near a finite endpoint and local ODE existence would extend it. Thus every restricted flow is complete.

Substituting the complete commuting restricted flows into the formula of step 1.1 defines a global ${\mathbb{R}}^n$-action on the compact fibre. Without compactness, only the local action is asserted.